Momentum conservation


Newton's third law tells us that when two objects interact, they exert forces on each other and those forces are equal and opposite:

$$  \overrightarrow{F}^{type}_{A \rightarrow B} = - \overrightarrow{F}^{type}_{B \rightarrow A}$$

for any type of force. When we combine this with the momentum form of N2, something interesting happens. We get one of the most powerful principles in all of physics: the law of momentum conservation. Let's see how this comes about by doing what we often do in physics. We consider the simplest possible case — a toy model.

A toy model of interactions - two objects

While any real world situation typically has lots of objects involved, we can get insight into how momentum conservation happens by considering what happens with just two interaction objects.

Suppose that we have two objects, A and B, and they interact with nothing else except each other (with an unspecified type of force). Then Newton's second law for the two objects become:

$$\frac{d \overrightarrow{p}_A}{dt} = \overrightarrow{F}_{B \rightarrow A}$$

$$\frac{d \overrightarrow{p}_B}{dt} = \overrightarrow{F}_{A \rightarrow B}$$

We've put the momenta on the left because that's what we want to pay attention to. Now let's do something a little different. Let's consider the AB pair of objects as a single system by adding these two equations together. (Actually, remember that adding two different objects together and treating them as a system played a role in our deciding that Newton's third law might actually be a reasonable thing to assume and test experimentally.) The result is

$$\frac{d (\overrightarrow{p}_A + \overrightarrow{p}_B)}{dt} =  \overrightarrow{F}_{B \rightarrow A} + \overrightarrow{F}_{A \rightarrow B}$$

But since the two forces are equal and opposite -- whatever kind of force they are -- by N3, this cancels!  The result is interesting:

$$\frac{d (\overrightarrow{p}_A + \overrightarrow{p}_B)}{dt} =  0$$

So since the sum of the momenta don't change, we get that

$\overrightarrow{p}_A + \overrightarrow{p}_B =$ constant

Now you might say, "So what?" But when something has its derivative equal to 0 that means that it doesn't depend on the variable you are differentiating with respect to, in this case time. This says that the total momentum of the system doesn't change with time.  But the individual momenta might change dramatically! If the two objects are billiard balls that are colliding they might take off in totally different directions.  Each momentum changes a lot, but the total doesn't!

What's happening physically?

The manipulations above are "just math" — interesting, but can we think about what those equations are telling us in a more physical way so we can see the mechanism of how this constancy arises? 

The way to do that is to let our "dt" become a small time interval "$\Delta t$| and us the Impulse-Momentum form of N2. First we change our derivative form to a delta form:

$$\frac{\Delta \overrightarrow{p}_A}{\Delta t} = \overrightarrow{F}_{B \rightarrow A}$$

$$\frac{\Delta \overrightarrow{p}_B}{\Delta t} = \overrightarrow{F}_{A \rightarrow B}$$

Then we multiply both sides by $\Delta t$ to get the Impulse-Momentum form:

$$\Delta \overrightarrow{p}_A = \overrightarrow{F}_{B \rightarrow A} \Delta t$$

$$\Delta \overrightarrow{p}_B = \overrightarrow{F}_{A \rightarrow B} \Delta t$$

Since the two forces are equal and opposite by N3, this form tells us that

When two objects interact, the change in momentum in the first caused by the second is equal and opposite to the change in momentum in the second caused by the first.

So when two objects interact, the are not creating momentum, but exchanging it. (Since momentum is a vector, they could both start at 0 and repel and start moving apart, one getting positive momentum and the other negative momentum.) 

A more realistic analysis

Let's try to make this a little more general. Suppose we have two objects, A and B, that we are going to consider as a "system".  Each will interact with each other by some (unspecified type of) force and with other objects in the world. We'll refer to those objects that are not A or B as "external" to the system.  Then our N2 equations (in momentum form) become

$$\frac{d \overrightarrow{p}_A}{dt} = \overrightarrow{F}_{B \rightarrow A} + \overrightarrow{F}^{ext}_A$$

$$\frac{d \overrightarrow{p}_B}{dt} = \overrightarrow{F}_{A \rightarrow B} + \overrightarrow{F}^{ext}_B$$

If we add these two together, taking into account that the forces that A and B exert on each other will cancel, we get the result

$$\frac{d \overrightarrow{p}_A}{dt} + \frac{d \overrightarrow{p}_B}{dt} = \overrightarrow{F}^{ext}_A + \overrightarrow{F}^{ext}_B = \overrightarrow{F}^{ext}_{AB}$$

We can now easily see the conditions for us to get the "total momentum of the system doesn't change" result again.

If we have a system of two interacting objects such that the net external force on the objects cancel, then the total momentum of the system doesn't change even though the individual momentum might change.

We refer to this as the Theorem of Momentum Conservation.  We can easily see that this can be generalized to any number of objects since for every pair, N3 will say that they will cancel when you add the N2 equations together. The general result is

If we have a system consisting of any number of interacting objects such that the net external force on the objects cancel, then the total momentum of the system doesn't change even though the individual momenta might.

It's interesting to realize that whether momentum conservation holds depends on what system we are considering. If we only consider the thrown ball, momentum is not conserved for the ball because there is an external force — gravity. But if we included the earth (and if we were able to measure its momentum to an incredible accuracy) then the momentum of the two would be conserved. 

Total vs change

There are a number of different ways that momentum conservation can be expressed.  Let's just go back to our toy model and consider two objects for whom the net external forces cancel. We then have the result

$$\frac{d (\overrightarrow{p}_A + \overrightarrow{p}_B)}{dt} =  0$$

$\overrightarrow{p}_A + \overrightarrow{p}_B =$ constant

We can express this in a variety of ways.

We can say that if we look at the total momenta at two different times the result for the total is the same. Or we could say that the change is 0. 

For two items this becomes particularly useful since the total of the two stays the same says that they must change in equal but opposite ways.

$\overrightarrow{p}_A + \overrightarrow{p}_B =$ constant

$$\Rightarrow \quad \overrightarrow{p}^i_A + \overrightarrow{p}^i_B =  \overrightarrow{p}^f_A + \overrightarrow{p}^f_B$$

that is, the sum of the initial momenta is equal to the sum of the final momenta.

We can also look at it in terms of the change:

$\overrightarrow{p}_A + \overrightarrow{p}_B =$ constant

$$\Rightarrow \quad \Delta (\overrightarrow{p}_A + \overrightarrow{p}_B) = 0$$

$$\Rightarrow \quad \Delta \overrightarrow{p}_A = - \Delta\overrightarrow{p}_B$$

So there are two ways we can express the fact that the momentum of a pair of objects is conserved:

  • The momenta added together at any (initial) time is equal to the sum of the momenta at a later (final) time.
  • That change in one object's momentum over any time interval is equal and opposite to the change in the other object's momentum.

Be careful! This is a place where if you are sloppy or are doing one-step thinking, you can easily mess up. There are examples we will do in which one of the objects has momentum 0 either initially or finally and for that special case the momentum and the change in the momentum are the same.  But that is not usually the case! Be very careful to distinguish the momentum from its change and to state your momentum conservation carefully, paying attention to whether you are talking about a momentum or a change in momentum.

Workout: Momentum conservation


Joe Redish 10/18/11


Article 398
Last Modified: October 16, 2019